Boundary value problem with substitution

[TheFerruccio]

Homework Statement

Find the general solution to the boundary value problem.

Homework Equations

$$(xy')' + \lambda x^{-1}y = 0$$
$$y(1) = 0$$
$$y(e) = 0$$
use $$x = e^t$$

The Attempt at a Solution

$$x = e^t$$ so $$\frac{dx}{dt} = e^t$$

using chain rule:
$$y' = e^{-t}\frac{dy}{dt}$$

Substituting this in:

$$\frac{d}{dx}(e^t(e^{-t}\frac{dy}{dt})) + \lambda e^{-t}y = 0$$
$$\frac{d}{dx}(\frac{dy}{dt}) = \lambda e^{-t}y = 0$$
$$\frac{d}{dx}(y'e^t) + \lambda e^{-t}y = 0$$

From this point, I feel like I am going in circles. I want to get everything in equal powers of $$e$$, so I can cancel it out, get a general solution in terms of $$t$$, and plug in the boundary values to find my constants.

 

[hunt_mat]

Expand out the ODE fully:
$$\frac{dy}{dx}+x\frac{d^{2}y}{dx^{2}}+\frac{\lambda}{x}y\Rightarrow x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}+\lambda y=0$$
This is called the Euler equation, look for solutions of the form x^{n}.

 

[TheFerruccio]

Expand out the ODE fully:
$$\frac{dy}{dx}+x\frac{d^{2}y}{dx^{2}}+\frac{\lambda}{x}y\Rightarrow x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}+\lambda y=0$$
This is called the Euler equation, look for solutions of the form x^{n}.

How did you arrive at that?

Here is my attempt at expanding it out fully:

$$\frac{d}{dx}(x\frac{dy}{dx}) + \frac{\lambda}{x}y = 0$$
$$\frac{dy}{dx} + x\frac{d^2y}{dx^2} + \frac{\lambda}{x}y = 0$$
$$x^2y'' + y'x + \lambda y = 0$$

... Ah, I see what you did, ok. Thanks for the tip. Do I even need a substitution with this?

 

[TheFerruccio]

I continued the problem, using the method you suggested, and I ended up with the following differential equation, in terms of t:

$$\ddot{y} + \lambda y = 0$$

The general solution to this, in terms of $$t$$, is:

$$y = A\cos{\sqrt{\lambda}t} + B\sin{\sqrt{\lambda}t}$$

Boundary conditions holding true, I get:

$$A = 0$$
$$B$$ is arbitrary

$$y(x) = Bsin(n\ln{x})$$ where $$n = 1, 2, 3, \ldots$$

If you want, I can be more thorough as to how I got this answer, especially if it's wrong!

 

[hunt_mat]

As I said sibstitute in y=x^{n} to obtain the following:
$$\left[ n(n-1)+n+\lambda\right] x^{n}=0$$
We require an equation for n, just like the constant coefficient case, can you say what this equation is?

I think that in general you're over thinking it. You don't need a substitution for this equation.

 

[TheFerruccio]

Using the substitution method, when solving using the second boundary condition, I am ending up with a discrepancy when attempting a solution with both methods, and it's minor.

For the method of substituting like what the problem suggests, I get:

$$0 = B\sin{\sqrt{\lambda}}$$

So...

$$\lambda = n^2\pi^2$$

Using the euler format, I am getting...

$$0 = B\sin{n\sqrt{\lambda}}$$

So...

$$\lambda = \frac{m^2pi^2}{n^2}$$

where $$m$$, in this case, is $$n$$ in the first method.

I checked my algebra, and I don't see how these can be so different. Is it because $$n$$ and $$m$$ are arbitrary, and can combine to equal a single $$n$$?

 

[hunt_mat]

The two solutions are:
$$n=\pm\sqrt{\lambda}$$
So the solution becomes:
$$y=Ax^{\sqrt{\lambda}}+Bx^{-\sqrt{\lambda}}$$
Then just use the boundary conditions to find A and B.

 

[TheFerruccio]

The two solutions are:
$$n=\pm\sqrt{\lambda}$$
So the solution becomes:
$$y=Ax^{\sqrt{\lambda}}+Bx^{-\sqrt{\lambda}}$$
Then just use the boundary conditions to find A and B.

Really? So, the solution is not periodic at all? That's strange. I kept getting imaginary roots, which, by doing it two different methods, resulted in getting a sine and cosine. I will post my work in a little bit.

 

[hunt_mat]

I made a mistake in my algebra, my solutions should have been
$$n=\pm i\sqrt{\lambda}$$
Then my solution reduces to yours... The solution my method becomes:
$$y=\sum_{n=0}^{\infty}A_{n}\left( x^{n\pi i}-x^{-n\pi i}\right)$$

 

[TheFerruccio]

I made a mistake in my algebra, my solutions should have been
$$n=\pm i\sqrt{\lambda}$$
Then my solution reduces to yours... The solution my method becomes:
$$y=\sum_{n=0}^{\infty}A_{n}\left( x^{n\pi i}-x^{-n\pi i}\right)$$

That's actually an interesting notation. It seems to be more properly descriptive of what all the solutions are.

 

[hunt_mat]

By writing
$$x^{k}=e^{k\log x}$$
My solutions become
$$y=\sum_{n=0}^{\infty}B_{n}\sin (n\pi\log x)$$